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Describe the region of R^3, sphere with inequality

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Describe the region of R^3 that is represented by:

    2. Relevant equations

    x^2 + y^2 + z^2 > 2z


    3. The attempt at a solution

    I'm not sure what to do with this at, especially at z=0 and z=2
     
  2. jcsd
  3. Sep 23, 2011 #2
    Maybe you should complete the square
     
  4. Sep 23, 2011 #3
    x^2 + y^2 + (z-1)^2 > 1
    a sphere with centre (0,0,1) and radius >1?
     
  5. Sep 23, 2011 #4

    SammyS

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    Seems like that could be many spheres.
     
  6. Sep 23, 2011 #5
    would it be:
    x^2 + y^2 + (z-1)^2 > 1

    infinite number of spheres with centre (0,0,1) and radius >1?
     
  7. Sep 23, 2011 #6

    SammyS

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    Where do you find any point (x, y) that satisfies x^2 + y^2 + (z-1)^2 > 1 , relative to the set of points that satisfy x^2 + y^2 + (z-1)^2 = 1 ?
     
  8. Sep 24, 2011 #7
    I'm lost now. I am fairly sure I have the centre correct at (0,0,1), the only thing I can think of regarding your previous post is:
    (x,y) = (x>1, y>1) or
    (x,y) = (x>=1, y>=1). But this seems incorrect.
     
  9. Sep 24, 2011 #8

    SammyS

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    Yes, the equation x2 + y2 + (z-1)2 = 1 describes a sphere of radius 1 with center at (x, y, z) = (0, 0, 1) .

    Pick any point that is outside of the sphere. What do you get for x2 + y2 + (z-1)2 for that point?

    Pick any point that is inside of the sphere. What do you get for x2 + y2 + (z-1)2 for this point?
     
  10. Sep 25, 2011 #9
    for the equation, a point inside the sphere would be (1/2, 0, 1)
    .5^2 + 0^2 + (1-1)^2 = .25

    outside the the sphere would be (1,1,1)
    1^2 + 1^2 + (1-1)^2 = 2
     
  11. Sep 25, 2011 #10

    SammyS

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    Notice that .25 < 1, so for point (1/2, 0, 1), x2 + y2 +(z-1)2 < 1 .


    Similarly, 2 > 1, , so for point (1, 1, 1), x2 + y2 +(z-1)2 > 1 .

    Does that give you any ideas?

    BTW, what does [itex]\sqrt{(x-0)^2 + (y-0)^2 + (z-1)^2}[/itex] represent ?
     
  12. Sep 25, 2011 #11
    \sqrt{(x-0)^2 + (y-0)^2 + (z-1)^2}
    is the distance between two points, (0, 0, -1) and any point (x,y,z)
    but I still don't understand the constraint on x, y, and/or z.
     
  13. Sep 25, 2011 #12
    or the inequality of the radius
     
  14. Sep 25, 2011 #13

    HallsofIvy

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    I'm afraid SammyS rather confused things when he asked about "(x, y)" in post #6.

    alias, In post #9, you said, "outside the the sphere would be (1,1,1)". Yes, that is a point in this set but any point on that set lies in this set. The only problem with your post #5, "infinite number of spheres with centre (0,0,1) and radius >1?", is not enough. The set of all spheres with integer radius, greater than 1, satisfies that but does not include points that satisfy, for example, [itex]x^2+ y^2= \frac{9}{4}[/itex].
     
  15. Sep 25, 2011 #14

    SammyS

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    Yes. TYPO !!!

    Sorry about that.

    I meant to type " point (x, y, z) = ... "
     
  16. Sep 25, 2011 #15

    SammyS

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    Actually, it's the distance between two points, (0, 0, 1) and any point (x,y,z). In other words, it's the distance from the center of your sphere to any point (x,y,z).

    If [itex]\sqrt{(x-0)^2 + (y-0)^2 + (z-1)^2}>1\,, [/itex] where is the point (x, y, z) located in relation to the sphere ?
     
  17. Sep 26, 2011 #16
    The distance between (0,0,1) and (x,y,z) is greater than 1. I'm not sure about the relation to the sphere though.
     
  18. Sep 26, 2011 #17

    SammyS

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    What is the distance from (0,0,1) to any point on the surface of the sphere ?
     
  19. Sep 26, 2011 #18
    The distance from (0,0,1) to any point on the surface of the sphere is 1.
    How does this relate to the inequality with the radius?
     
  20. Sep 26, 2011 #19

    SammyS

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    If the distance from (0,0,1) is 1, the point (x,y,z) is on the surface of the sphere. If that distance is greater than 1 doesn't the point lies outside the sphere?

    If [itex](x)^2 + (y)^2 + (z-1)^2>1\,,[/itex] what does that say about [itex]\sqrt{(x-0)^2 + (y-0)^2 + (z-1)^2}\,?[/itex]
     
  21. Sep 26, 2011 #20
    Yes, any point (x,y,z) that has a distance greater than 1 from (0,0,1) lies outside the the sphere. So the original eqn, x^2 + y^2 + (z-1)^2 > 1 consists of all points outside the outside the sphere with with centre (0,0,1) and radius 1? Would that not be all spheres with radius >1 and the same centre?
     
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